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82                           HlSTOBY OP THE THEORY OF NUMBERS.                 [CHAP. Ill
where \k\ =p(p—l) . . .(p—K). Hence, if p is a prime and n<p~ 1, sn+i— sn=0. But $i=0. Hence sn=0(n<p — 1), sp..i= — (p— 1)1 Thus Wilsonjs theorem follows from Fermat's.
Without giving references, Aubry (p. 298) attributed Homer's37 proof of Euler's theorem to Gauss; the proof (pp. 439-440) by Paoli46 (and, Thue115) of Fermat's theorem to Euler12; the proof (p. 458) by Laplace23 of Euler's theorem by powering to Euler.
R. D. Carmichael138 noted that, if L is the 1. c. m. of all the roots z of <t>(z)=a, and if x is prime to I/, then xa^l (mod L). Hence except when n and n/2 are the only numbers whose ^-function is the same as that of n, a^(n>==l holds for a modulus M which is some multiple of n. A practical method of finding M is given.
R. D. Carmichael139 proved the first result by Lucas.110
J. A, Donaldson140 deduced Fermat's theorem from the theory of periodic fractions.
W. A. Lindsay141 proved Fermat's theorem by use of the binomial theorem.
. J. I. Tschistjakov142 extended Euler's theorem as had Lucas.110
P. Bachmann143 proved the remarks by Lucas,110 but replaced 0+(r<n by n^(t>-\-ff} stating that the sign is > if n is divisible by at least two distinct primes.
A. Thue144 noted that a different kinds of objects can be placed into n given places in an ways. Of these let C7J be the number of placings such that each is converted into itself by not fewer than n applications of the operation which replaces each by the next and the last by the first. Then VI is divisible by n. If n is a prime, Una = an— a and we have Fermat's theorem. Next, an=^Uda} where d ranges over the divisors of n. Finally, if p, q, . . . , r are the distinct prune factors of n,
where D ranges over the distinct divisors of pq. . .r, while 6 is the number of prime factors of D. Euler's theorem is deduced from this.
H. C. Pocklington145 repeated Bricard's131 proof.
U. Scarpis146 proved the generalized Wilson theorem by a method similar to Arndt's.70 The case of modulus 2X (X>2) is treated by induction. Assume that IIr=l (mod 2X), where rlf . . ., rv are the 0=<£(2X) odd integers <2\ Then rlf. . ., rr, r!+2x,. . ., rp+2x are the residues modulo 2X+1 and their product is seen to be ^=1 (mod 2M~1). Next, let the modulus be
"'Bull. Amer. Math. Soc., 15, 1908-9, 221-2.
"TOd., 16, 1909-10, 232-3.
140Edinburgh Math. Soc. Notes, 1909-11, 79-84.
M17bid., 78-79.
M2Tagbl. XII Vers. RUBS. Nat., 124, 1910 (Russian),
mNiedere Zahlentheorie, II, 1910, 43-44.
144Skrifter Videnskabs-Selskabet, Christiania, 1910, No. 3, 7 pp.
^Nature, 84, 1910, 531.
"•Periodico di Mat., 27, 1912, 231-3.